POJ 1730 Perfect Pth Powers
2016-04-07 12:37
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Perfect Pth Powers
Description
We say that x is a perfect square if, for some integer b, x = b2. Similarly, x is a perfect cube if, for some integer b, x = b3. More generally, x is a perfect pth power
if, for some integer b, x = bp. Given an integer x you are to determine the largest p such that x is a perfect pth power.
Input
Each test case is given by a line of input containing x. The value of x will have magnitude at least 2 and be within the range of a (32-bit) int in C, C++, and Java. A line containing 0 follows the last test case.
Output
For each test case, output a line giving the largest integer p such that x is a perfect pth power.
Sample Input
Sample Output
题目大意:要求出一个完美的平方数,完美的平方数是这样的,n = bp的当P最大的时候才是一个完美的平方数
解题思路:将P从31到1遍历枚举,使用POW函数将他求出来,不过有一点问题就是精度问题,当用POW(125,1/3),直接取整的时候是4,所以需要在后面加一个0.1,也就是(int)(POW(125,1/3)+0.1).这样就可以求出结果了,还有要注意的是,他给出的N可能是负数,所以要对负数特殊处理,也就是说当N是负数的时候,P不可能是偶数,只能是奇数。
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 17339 | Accepted: 3974 |
We say that x is a perfect square if, for some integer b, x = b2. Similarly, x is a perfect cube if, for some integer b, x = b3. More generally, x is a perfect pth power
if, for some integer b, x = bp. Given an integer x you are to determine the largest p such that x is a perfect pth power.
Input
Each test case is given by a line of input containing x. The value of x will have magnitude at least 2 and be within the range of a (32-bit) int in C, C++, and Java. A line containing 0 follows the last test case.
Output
For each test case, output a line giving the largest integer p such that x is a perfect pth power.
Sample Input
17 1073741824 25 0
Sample Output
1 30 2
题目大意:要求出一个完美的平方数,完美的平方数是这样的,n = bp的当P最大的时候才是一个完美的平方数
解题思路:将P从31到1遍历枚举,使用POW函数将他求出来,不过有一点问题就是精度问题,当用POW(125,1/3),直接取整的时候是4,所以需要在后面加一个0.1,也就是(int)(POW(125,1/3)+0.1).这样就可以求出结果了,还有要注意的是,他给出的N可能是负数,所以要对负数特殊处理,也就是说当N是负数的时候,P不可能是偶数,只能是奇数。
#include <cstdio> #include <cmath> using namespace std; //typedef long long ll; //ll x; int x; int main() { while (scanf("%d", &x), x){ if (x > 0){ for (int i = 31; i>= 1; i--){ int a = (int)(pow(x * 1.0, 1.0 / i) + 0.1); int b = (int)(pow(a * 1.0, 1.0 * i) + 0.1); if (x == b){ printf("%d\n", i); break; } } } else{ x = -x; for (int i = 31; i>= 1; i -= 2){ int a = (int)(pow(x * 1.0, 1.0 / i) + 0.1); int b = (int)(pow(a * 1.0, 1.0 * i) + 0.1); if (x == b){ printf("%d\n", i); break; } } } } return 0; }
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